Fluid Mechanics Viscosity Question 253
Question: The force acting on a window of area$ 50cm\times 50cm $ of a submarine at a depth of 2000 m ill an ocean, interior of which is maintained at sea level atmospheric pressure is (Density of sea water$ =10^{3}kg{{m}^{-3}},g=10m{{s}^{-2}} $ )
Options:
A) $ 10^{6}N $
B) $ 5\times 10^{5}N $
C) $ 25\times 10^{6}N $
D) $ 5\times 10^{6}N $
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Answer:
Correct Answer: D
Solution:
[d] The pressure outside the submarine is $ P=P_{a}+\rho gh $ Pressure inside the submarine is $ P_{a} $ . Net pressure acting on the window is $ P_{g}=P-P_{a}=\rho gh $ $ {{10}^{-3}}kg{{m}^{-3}}\times 10m{{s}^{-2}}\times 2000m $ $ =2\times 10^{7}Pa $ Area of window is $ A=50cm\times 50cm=2500\times {{10}^{-4}}m^{2} $ Force on the window is $ F=P_{g}A=2\times 10^{7}Pa\times 2500\times {{10}^{-4}}m^{2} $ $ =5\times 10^{6}N $
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